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2 years ago

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The strength of the electric field 0.5 m from a 6 µc charge is n/c. (use k = 8.99 × 109 n•meters squared per coulomb squared and

round answer to the nearest whole number.)

1 answer:

diamong [38]2 years ago

6 0

53....................................

Explanation:

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What is the diameter of a 12lb shot if the specific gravity is of the shot iron in the shot is 6.8, the density of fresh water 6

marysya [2.9K]

Answer:

The diameter is 0.378 ft.

Explanation:

Given that,

Mass of shot = 12 lb

Density of fresh water = 62.4 lb/ft

Specific gravity = 6.8

We need to calculate the volume of shot

$V = \dfrac{4}{3}\pi r^3\ ft^3$

The density of shot is

Using formula of density

$\rho = \dfrac{m}{V}$

Put the value into the formula

$\rho =\dfrac{12}{ \dfrac{4}{3}\pi r^3}$

We need to calculate the radius

Using formula of specific gravity

$specific\ gravity =\dfrac{density\ of\ shot}{dnsity\ of\ water}$

Put the value into the formula

$6.8=\dfrac{\dfrac{12}{\dfrac{4}{3}\pi r^3}}{62.4}$

$r^3=\dfrac{12}{\dfrac{4}{3}\pi\times6.8\times62.4}$

$r^3=0.0067514$

$r =(0.0067514)^{\frac{1}{3}}$

$r=0.1890\ ft$

The diameter will be

$d = 2\times r$

$d =2\times0.1890$

$d =0.378\ ft$

Hence, The diameter is 0.378 ft.

7 0

3 years ago

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

(c) In the first part we have:

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$

$t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s$

So, the total time is:

$t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s$

7 0

4 years ago

Kraig pulls a box to the right at an angle of 40 degrees to the horizontal with a force of 30 Newtons. If Kraig pulls the box a

eimsori [14]

Answer:

459.6J

Explanation:

Given parameters:

Angle of pull = 40°

Force applied = 30N

Distance moved = 20m

Unknown:

Work done by Kraig = ?

Solution:

To solve this problem;

Work done = F x dcosФ

d is the distance

F is the force

Ф is the angle given

Work done = 30 x 20cos40° = 459.6J

3 0

3 years ago

A 0.5 kg block slides down a frictionless semicircular track from a height, h, and collides with a second 1.5 kg block at the bo

Ede4ka [16]

Answer:

$h' = 0.062\cdot h$

Explanation:

The speed of the 0.5 kg block just before the collision is found by the Principle of Energy Conservation:

$U_{g} = K$

$m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot h}$

$v \approx 4.429\cdot \sqrt{h}$

Knowing that collision is inelastic, the speed just after the collision is determined with the help of the Principle of Momentum Conservation:

$(0.5\,kg) \cdot (4.429\cdot \sqrt{h}) = (2\,kg)\cdot v$

$v = 1.107\cdot \sqrt{h}$

Lastly, the height reached by the two blocks is:

$K = U_{g}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot h'$

$h' = \frac{v^{2}}{2\cdot g}$

$h' = \frac{(1.107\cdot \sqrt{h})^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h' = 0.062\cdot h$

3 0

3 years ago

A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field that incre

mihalych1998 [28]

Answer:

Explanation:

Area of crossection, A = 7.80 cm²

Initial magnetic field, B = 0.5 T

Final magnetic field, B' = 3.3 T

Time, t = 1 s

resistance of the coil, R = 1.2 ohm

The induced emf is given by

$e=\frac{d\phi}{dt}=A\frac{B' - B}{t}$

where, Ф is the rate of change of magnetic flux.

e = 7.80 x 10^-4 x (3.3 - 0.5) / 1

e = 2.184 mV

i = e/R

i = 2.184/1.2

i = 1.82 mA

4 0

3 years ago